Step-by-step explanation:
Properties of an Isosceles Triangle
(Most of this can be found in Chapter 1 of B&B.)
Definition: A triangle is isosceles if two if its sides are equal.
We want to prove the following properties of isosceles triangles.
Theorem: Let ABC be an isosceles triangle with AB = AC. Let M denote the midpoint of BC (i.e., M is the point on BC for which MB = MC). Then
a) Triangle ABM is congruent to triangle ACM.
b) Angle ABC = Angle ACB (base angles are equal)
c) Angle AMB = Angle AMC = right angle.
d) Angle BAM = angle CAM
Corollary: Consequently, from these facts and the definitions:
Ray AM is the angle bisector of angle BAC.
Line AM is the altitude of triangle ABC through A.
Line AM is the perpendicular bisector of B
Segment AM is the median of triangle ABC through A.
Proof #1 of Theorem (after B&B)
Let the angle bisector of BAC intersect segment BC at point D.
Since ray AD is the angle bisector, angle BAD = angle CAD.
The segment AD = AD = itself.
Also, AB = AC since the triangle is isosceles.
Thus, triangle BAD is congruent to CAD by SAS (side-angle-side).
This means that triangle BAD = triangle CAD, and corresponding sides and angles are equal, namely:
DB = DC,
angle ABD = angle ACD,
angle ADB = angle ADC.
(Proof of a). Since DB = DC, this means D = M by definition of the midpoint. Thus triangle ABM = triangle ACM.
(Proof of b) Since angle ABD = angle ABC (same angle) and also angle ACD = angle ACB, this implies angle ABC = angle ACB.
(Proof of c) From congruence of triangles, angle AMB = angle AMC. But by addition of angles, angle AMB + angle AMC = straight angle = 180 degrees. Thus 2 angle AMB = straight angle and angle AMB = right angle.
(Proof of d) Since D = M, the congruence angle BAM = angle CAM follows from the definition of D. (These are also corresponding angles in congruent triangles ABM and ACM.)
QED*
*Note: There is one point of this proof that needs a more careful “protractor axiom”. When we constructed the angle bisector of BAC, we assumed that this ray intersects segment BC. This can’t be quite deduced from the B&B form of the axioms. One of the axioms needs a little strengthening.
The other statements are immediate consequence of these relations and the definitions of angle bisector, altitude, perpendicular bisector, and median. (Look them up!)
Definition: We will call the special line AM the line of symmetry of the isosceles triangle. Thus we can construct AM as the line through A and the midpoint, or the angle bisector, or altitude or perpendicular bisector of BC. Shortly we will give a general definition of line of symmetry that applies to many kinds of figure.
Proof #2 (This is a slick use of SAS, not presented Monday. We may discuss in class Wednesday.)
The hypothesis of the theorem is that AB = AC. Also, AC = AB (!) and angle BAC = angle CAB (same angle). Thus triangle BAC is congruent to triangle BAC by SAS.
The corresponding angles and sides are equal, so the base angle ABC = angle ACB.
Let M be the midpoint of BC. By definition of midpoint, MB = MC. Also the equality of base angles gives angle ABM = angle ABC = angle ACB = angle ACM. Since we already are given BA = CA, this means that triangle ABM = triangle ACM by SAS.
From these congruent triangles then we conclude as before:
Angle BAM = angle CAM (so ray AM is the bisector of angle BAC)
Angle AMB = angle AMC = right angle (so line MA is the perpendicular bisector of BC and also the altitude of ABC through A)
QED
Faulty Proof #3. Can you find the hole in this proof?)
In triangle ABC, AB = AC. Let M be the midpoint and MA be the perpendicular bisector of BC.
Then angle BMA = angle CMA = right angle, since MA is perpendicular bisector.
MB = MC by definition of midpoint. (M is midpoint since MA is perpendicular bisector.)
AM = AM (self).
So triangle AMB = triangle AMC by SAS.
Then the other equal angles ABC = ACB and angle BAM = angle CAM follow from corresponding parts of congruent triangles. And the rest is as before.
QED??
